Re: Faraday induction

[Eugene Mosca <emosca@PTD.NET>]

On 6/1/01 11:40 AM, "John S. Denker" <jsd@MONMOUTH.COM> wrote:

> Eugene Mosca wrote:
> > > How about energy per unit charge?
> >
> > Does this mean that the kinetic energy per unit charge of charged particle
> > in a beam passing through a surface is voltage?
>
> No, that's not what I had in mind.
>
> Loosely speaking, I was thinking of the energy-per-unit-charge of an
> idealized test particle that had no kinetic energy, no gravitational
> potential energy, et cetera.
>
> More precisely, I was thinking of the part of the energy due solely to
> the interaction of the particle's charge with the applied field.  This
> could be ascertained by comparing the energy of the charged particle
> with the energy of an otherwise-similar uncharged particle.

I do not follow this.  What is the energy of the interaction of the particle
with the applied field?  If kinetic energy is neglected then it seems to me
the only other energy is electrostatic potential energy.  However, that is
covered under the potential concept and the voltage concept is not needed.


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